A uniform solid sphere of mass m and radius ‘r’ is released from rest from the top of a wedge of mass ‘2m’ initially at rest on a smooth horizontal surface as shown. There is sufficient friction between the sphere and the wedge so that the sphere rolls on the wedge without slipping. The sphere leaves the wedge horizontally at A after descending through a vertical displacement h. Let $V_{1}, V_{2} a n d ω$ be the magnitudes of the velocity of the centre of sphere, velocity of wedge and the angular velocity of the sphere at the instant the sphere leaves the wedge. Choose the correct statements.

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$V_{1} + V_{2} = r ω$

Total mechanical energy of sphere and wedge system is conserved during the process

The ratio of total kinetic energy of the sphere and to the wedge at the instant the sphere leaves the wedges is 19:5

$V_{1} = \sqrt{\frac{5 g h}{6}}$

answer is A, B, C, D.

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Detailed Solution

Condition for rolling:
$v_{1} + v_{2} = R w \Rightarrow w \equiv \frac{3 v_{1}}{2 R} C L M; m v_{1} = 2 m v_{2} \Rightarrow v_{2} = \frac{v_{1}}{2} k_{w e d g e} = \frac{1}{2} \times 2 m v_{2}^{2} = \frac{m v_{1}^{2}}{4} k_{s p h e r e} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} \times \frac{2}{5} m R^{2} \times {(\frac{3 v_{1}}{2 R})}^{2} = \frac{19}{20} m v_{1}^{2} ∴ R a t i o = \frac{19}{5} C O E : \frac{m v_{1}^{2}}{4} {1 + \frac{19}{5}} = m g h \Rightarrow v_{1} = \sqrt{\frac{5 g h}{6}}$